hdu1008 Elevator（模拟）

最新推荐文章于 2019-09-21 20:24:08 发布

原创最新推荐文章于 2019-09-21 20:24:08 发布 · 579 阅读

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#hdu1008 #模拟

本文介绍了一个简单的模拟过程问题，即如何计算电梯根据指定楼层请求列表完成所有任务所需的总时间。考虑了电梯上行和下行的不同速度及每层停留时间。

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Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 50934 Accepted Submission(s): 28102

Problem Description

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output

Print the total time on a single line for each test case.

Sample Input

Sample Output

  
   17
41

Author

ZHENG, Jianqiang

Source

ZJCPC2004

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解析：这就是一个简单的模拟过程。

电梯现在在 last，要去往 k 层，若上升，则花费时间为：6*（k-last）+5

若下降，则花费时间为：4*（last-k）+5

代码：

#include<cstdio>
using namespace std;

int main()
{
  freopen("1.in","r",stdin);
  int n,i,j,k,last,sum;
  while(scanf("%d",&n),n)
    {
      sum=0,last=0;
      for(i=1;i<=n;i++)
        {
          scanf("%d",&k);
          if(k>last)sum+=6*(k-last);
          else sum+=4*(last-k);
		  sum+=5,last=k;  
        }
      printf("%d\n",sum);  
    }
  return 0;
}